Question

The true dip of a normal limb associated with asymmetric folding is 20° and the angle that this limb makes with the axial planar cleavage is 35°. The true dip of the axial planar cleavage, in degree, is.........

Hint:

To calculate the true dip of the axial planar cleavage, use the relationship \( \tan(D_{\text{cleavage}}) = \tan(D_{\text{limb}}) \times \sin(\theta) \), where \( \theta \) is the angle between the limb and cleavage.

Answer 1

Correct Answer: 15

Given:
- The true dip of the normal limb \( D_{\text{limb}} = 20^\circ \)
- The angle between the limb and axial planar cleavage \( \theta = 35^\circ \)
Step 1: To find the true dip of the axial planar cleavage, we use the relationship between the dip of the limb and the cleavage:
\[ \tan(D_{\text{cleavage}}) = \tan(D_{\text{limb}}) \times \sin(\theta) \]

Step 2: Substitute the given values:
\[ \tan(D_{\text{cleavage}}) = \tan(20^\circ) \times \sin(35^\circ) \]

Step 3: Calculate the values:
\[ \tan(20^\circ) \approx 0.364 \] \[ \sin(35^\circ) \approx 0.574 \]
\[ \tan(D_{\text{cleavage}}) = 0.364 \times 0.574 = 0.209 \]

Step 4: Now, calculate the dip of the axial planar cleavage:
\[ D_{\text{cleavage}} = \tan^{-1}(0.209) \approx 15^\circ \]
Thus, the true dip of the axial planar cleavage is 15°.
\[ \boxed{15^\circ} \]